Write two possible quartic inequalities, one using the less than or equal to symbol and the other using the greater than or equal to symbol that
correspond to the following solution:
`-6-sqrt2 <= x <=-6+sqrt2` or `6-sqrt2<=x<=6+sqrt2`
To find the original inequalities that made these solutions, we need to rewrite each inequality as a quadratic inequality (since there are square roots in the solutions).
The roots of the first inequality are `-6-sqrt2` and `-6+sqrt2` . The roots can be combined to get:
`(x+6+sqrt2)(x+6-sqrt2)` expand out, noticing it's a difference of squares
`=(x+6)^2-2` now expand the brackets
`=x^2+12x+36-2` simplify the constants
We can see that the inequality `x^2+12x+34>=0` gives the first solution set above.
For the second inequality, we get the roots `6-sqrt2` and `6+sqrt2` . As in the earlier case, the roots can be combined to get:
`(x-6+sqrt2)(x-6-sqrt2)` expand noticing again difference of squares
`=(x-6)^2-2` expand brackets fully now
`=x^2-12x+36-2` simplify constants
Again, we see that the inequality `x^2-12x+34>=0` gives the second solution set.
Now combining the two quadratics will give the quartic inequality.
`(x^2+12x+34)(x^2-12x+34)>=0` rearrange to get difference of squares
`(x^2+34+12x)(x^2+34-12x)>=0` expand brackets
`(x^2+34)^2-(12x)^2>=0` expand brackets fully
The required quartic inequality is `x^4-76x^2+1156>=0` .
To get an inequality that uses the less than or equal to symbol, we need to multiply the polynomial by -1, to get: